Module 2. Static pressure of liquids

Lesson 2

STATIC PRESSURE OF LIQUIDS, HYDRAULIC PRESSURE, ABSOLUTE AND GAUGE PRESSURE

2.1 Introduction

Fluid has property that it exerts force on all the sides, top and bottom. Pressure exerted by fluid is given as force per unit area which is as follows:

The SI unit of pressure is newton per square metre (N/m²). This is also known as pascal (Pa). The values of standard atmospheric pressure are as follows:

· 760 mm of mercury column

· 10.3 m of water column

· 101.3 kN/m²

· 101.3 kPa

· 1 bar

2.2 Hydrostatic Law

Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container. The pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth in the fluid.

2.2.1 Pressure head of liquid

 Consider a fluid of density ρ, having fluid element of area ∆A at a depth distance ‘h’ from the top surface of liquid (Fig. 2.1).

Fig. 2.1 Forces acting on fluid element of height ∆h

Height of fluid element = ∆h

Pressure at top of fluid element = p

Force on the top of fluid element = p ∆A

Weight of fluid element = ∆A. ∆h. ρ g

Upward force acting at bottom of the fluid element =   ̶ ̶

Under equilibrium conditions (downward force = upward force),

Pressure variation in any static fluid is described by the above basic pressure height relationship. This equation describes hydrostatic law and indicates that rate of increase of pressure in a vertical downward direction is equal to the weight density (also known as specific weight) of fluid at that point.

Integrating,

P =

Here h is known as the pressure head and its unit is in metres (m).

2.3 Liquid Paradox

The pressure in a liquid is not a function of shape or size of the container. Pressure is only the function of density and height or depth ‘h’ inside the liquid at which pressure has to be calculated.(Fig. 2.2)

Q. 1: Calculate the pressure at 5 m, 10 m and 15 m in a tank filled with water.

Solution:    P₁ at 5 m =  = 1000 × 9.8 ×5 = 49 000 Pa

                      P₂ at 10 m =  = 1000 × 9.8 ×10 = 98 000 Pa

                      P₃ at 15 m =   = 1000 × 9.8 ×15 = 1 47 000 Pa

2.4 Gauge Pressure and Absolute Pressure

Gauge Pressure

It is convenient to measure pressure in terms of taking atmospheric pressure as reference datum. Pressure measured above atmospheric pressure is known as gauge pressure. The atmospheric pressure on the scale is marked as zero.

Absolute Pressure

Since, atmospheric pressure changes with atmospheric condition, a perfect vacuum is taken as an absolute standard of pressure. Pressure measured above perfect vacuum are called absolute pressure. The figure 2.3 explains the concept of gauge and absolute pressure.

Absolute pressure = atmospheric pressure + gauge pressure

P_abs = P_atm + P_gauge

Fig. 2.3 Concept of gauge and absolute pressure

2.5 Pascal’s Law

It states that the intensity of pressure at any point in a liquid at rest is same in all direction. We consider a fluid element of dimensions as shown in the figure 2.4. P_s is the pressure exerted on the inclined surface, δ is the linear dimension of the surfaces x, y and z.

Fig. 2.4 Fluid element

Please see (Fig. 2.5)

Equation for horizontal forces:

P_x.δ_y.δ_z = P_sSinϴ.δ_z.δ_s

P_x.δ_y.δ_z = δ_z.δ_s.P_s.

P_x = P_s ---------------(i)

Equation for vertical forces:

P_y.δ_x.δ_z = P_s.Cosθ.δ_s.δ_z + .δ_y.δ_x.δ_z.ρg

0

P_y.δ_x.δ_z = P_s.δ_s.δ_z + .δ_y.δ_x.δ_z.ρg

  Can be neglected as δ is very small

P_y = P_s  --------------- (ii)

From equation (i) and (ii) we have,

P_x = P_y = P_s

2.6 Numerical

Q1. A hydraulic press has a diameter ratio between the two piston of 8:1. The diameter of the larger piston is 600 mm and it is required to support a mass of 3500 kg. The press filled with hydraulic fluid of sp.gravity 0.8. Calculate the force required on the smaller piston to provide the required force: a) When the two pistons are at the same level. B) When the smaller piston is 2.6 m below the larger piston.

Solution.

Q2. The diameter of the piston of a hydraulic jack is 6 times greater than the diameter of the plunger which is 200 mm. They are both at same height and made of same material. The piston weighs 100 N and supports a mass of 50 kg. The jack is filled with oil of density 900 kg/m³. Calculate the force required at the plunger to support the piston and mass if carries at a level 10 cm above that of plunger.

Solution:

Q3. The ratio of the diameter of plunger and ram is 1:5, the diameter of plunger is 40 mm; find the weight lifted by the hydraulic press when the force is applied at plunger is 500 N.

Solution.    P₁ = P₂